Employment Planning Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1158 Description A project manager wants to determine the number of the workers needed in every month. He does know the minima…

Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4782    Accepted Submission(s): 2019 Problem Description A project manager wants to determine the number of the workers needed…

题目传送门 Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6242    Accepted Submission(s): 2710 Problem Description A project manager wants to determine the number of the workers…

Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7097    Accepted Submission(s): 3034 Problem Description A project manager wants to determine the number of the workers needed…

Employment Planning Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5292 Accepted Submission(s): 2262 Problem DescriptionA project manager wants to determine the number of the workers needed in eve…

Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5151    Accepted Submission(s): 2208 Problem Description A project manager wants to determine the number of the workers neede…

Employment Planning 有n个月,每个月有一个最小需要的工人数量\(a_i\),雇佣一个工人的费用为\(h\),开除一个工人的费用为\(f\),薪水为\(s\),询问满足这n个月正常工作的最小费用,\(n\leq 12\). 解 显然可以猜一个结论,因为工人数不确定,猜测每一个月的工人数量必然为某一个月的工人的最小数量,于是可以设\(f[i][j]\)表示前i个月,拥有工人数量\(b_j\)的最小费用,其中\(b\)为\(a\)的离散化数组,因此有 \(f[i][j]=min(m…

Problem地址:http://acm.hdu.edu.cn/showproblem.php?pid=1158 一道dp题,或许是我对dp的理解的还不够,看了题解才做出来,要加油了. 只能先上代码了. #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int MAXN = 100 + 28; int dp[15][MAXN]; int workers[…

又一次看题解. 万事开头难,我想DP也是这样的. 呵呵,不过还是有进步的. 比如说我一开始也是打算用dp[i][j]表示第i个月份雇j个员工的最低花费,不过后面的思路就完全错了.. 不过这里还有个问题,这样开数组j开多大比较好,难道要我开2^31-1这么大? 题解里面开了1000多,也许再小一点也能过吧. 因为有可能解雇一个人的花费比较大,所以某个月可能继续雇佣他这样总的算来是最省的. 所以第i个月可能雇佣的人数是从num[i] ~ NumMax. 首先对第一个月的费用初始化,就是(雇佣+薪水)…

题意:给出n个月,雇佣一个人所需的钱hire,一个人工作一个月所需要的钱salary,解雇一个人所需要的钱fire,再给出这n个月每月1至少有num[i]个人完成工作,问完成整个工作所花费的最少的钱是多少. 用dp[i][j]表示在第i个月雇佣j个人所需要的最少花费 先考虑只解雇人和聘请人的情况 for(j=num[i];j<=sum;j++) { ])//说明雇佣了人 dp[i][j]=dp[i-][num[i-]]+j*salary+(j-num[i-])*hire; else//说明解聘了…